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Sagot :
Answer:
The solution can be defined as follows:
Explanation:
Calculating the magnetic force:
[tex]\to \overrightarrow{F_B}=ILB \sin \beta\\\\[/tex]
Calculating the length:
[tex]\to L=\sqrt{(1.39)^2+(0.85)^2}=1.63\ m\\\\[/tex]
[tex]\tan \theta=\frac{0.85}{1.39}\\\\\theta=\tan^{-1} (\frac{0.85}{1.39})\\\\=31.45^{\circ}\\\\\beta =60+\theta\\\\=60+31.45\\\\= 91.45^{\circ}\\\\ \overrightarrow{F_B} =(36.6\times 10^{-3}) (1.63) (48.8\times 10^{-6})\sin 91.45^{\circ}\\\\=2.91 \times 10^{-6}\ N\\\\=2.91 \ \mu \ N[/tex]
Find the graph file in the attachment.

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